答案：
【解答】解: (I) $\because A B$ 为 $\odot O$ 的直径,
$$
\begin{aligned}
& \therefore \angle A C B=90^{\circ}, \\
& \because C \text { 为 } \widehat{\mathrm{B}} \text { 的中点, } \\
& \therefore \widehat{\mathrm{AC}}=\widehat{\mathrm{BC}}, \\
& \therefore \angle C A B=\angle C B A=45^{\circ}, \\
& \therefore A C=A B \cdot \cos \angle C A B=4 \sqrt{2} \text {; } \\
& \text { (II) } \because D F \text { 是 } \odot O \text { 的切线, } \\
& \therefore O D \perp D F, \\
& \because O D \perp B C, \angle F C B=90^{\circ}, \\
& \therefore \text { 四边形 } F C E D \text { 为矩形, } \\
& \therefore F D=E C,
\end{aligned}
$$
在 Rt $\triangle A B C$ 中 中, $\angle A C B=90^{\circ}, A C=2, A B=8$, 则 $B C=\sqrt{\mathrm{AB}^2-\mathrm{AC}}{ }^2=2 \sqrt{15}$,
$$
\begin{aligned}
& \because O D \perp B C, \\
& \therefore E C=\frac{1}{2} B C=\sqrt{15}, \\
& \therefore F D=\sqrt{15} .
\end{aligned}
$$